[HollyD]

I need to know how to work this problem:


Simplify.

3+i
_____
2-3i

[asterisk_man]

I was reminded how to solve this problem here: http://www.answers.com/topic/complex-number

Remember that for complex numbers (a+bi)*(a-bi)=(a^2+b^2) and (a^2+b^2) is not a complex number. So if we multiply the denominator by 2+3i we will no longer have a complex denominator... which will make the equation "simplier" :)

But if we just multiply the denominator by 2+3i we will have changed the value of the initial equation.
If you multiply the numerator and denominator both by (2+3i) you will be multiplying the equation by (2+3i)/(2+3i) which =1 so multiplying by this number does not change the value you start with.
Now we have (3+i)(2+3i)/((2-3i)(2+3i))

Simplifying the denominator with the rule given above gives (3+i)(2+3i)/(4+9)=(3+i)(2+3i)/13

Now we simply multiply the two terms in the numerator together. I don't know what method you learned for doing this, I learned to multiply First, Outer, Inner, Last. So I would expand (3+i)(2+3i) = 3*2+3*3i+i*2+i*3i which = 6+9i+2i-3 (remember that I*i=-1) then combine like terms. (6-3)+(9i+2i)=3+11i so (3+i)(2+3i) = 3+11i so our final result is
(3+11i)/13

I hope this helps you understand how to work through this problem and ones similar to it. Please go over everything I wrote and ask any questions you have.

[sammyp43]

decide if the given formula is recursive or explicit. Then find the third term.
an=3n(n+1)

[asterisk_man]

sammy, this is a different question so next time you should create a new question instead of adding it to an existing post.

I know it's hard to type equations here so is your equation:

or